Question Number 151560 by peter frank last updated on 21/Aug/21
$$\mathrm{show}\:\mathrm{that}\:\:\mathrm{i}^{\mathrm{i}} \:\:\mathrm{is}\:\mathrm{always}\:\mathrm{real} \\ $$
Answered by puissant last updated on 21/Aug/21
$${i}^{\mathrm{2}} =−\mathrm{1}\:\Rightarrow\:{i}=\sqrt{−\mathrm{1}} \\ $$$${i}^{{i}} =\:\left(\sqrt{−\mathrm{1}}\right)^{{i}} ={e}^{{iln}\left(\sqrt{−\mathrm{1}}\right)} ={e}^{\frac{{i}}{\mathrm{2}}{lni}^{\mathrm{2}} } =\:{e}^{{ilni}} \\ $$$$={e}^{{i}\frac{\pi}{\mathrm{2}}{i}} ={e}^{−\frac{\pi}{\mathrm{2}}} \:\in\:\mathbb{R}.. \\ $$
Commented by peter frank last updated on 22/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Olaf_Thorendsen last updated on 21/Aug/21
$${z}\:=\:{i}^{{i}} \\ $$$$\mathrm{ln}{z}\:=\:{i}\mathrm{ln}{i}\:=\:{i}\mathrm{ln}\left({e}^{{i}\frac{\pi}{\mathrm{2}}} \right)\:=\:{i}\left({i}\frac{\pi}{\mathrm{2}}\right)\:=\:−\frac{\pi}{\mathrm{2}} \\ $$$${z}\:=\:{e}^{−\frac{\pi}{\mathrm{2}}} \:\in\mathbb{R}. \\ $$
Commented by peter frank last updated on 22/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by MJS_new last updated on 21/Aug/21
$$\mathrm{i}^{\mathrm{i}} =\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \right)^{\mathrm{i}} =\mathrm{e}^{\mathrm{i}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}} =\mathrm{e}^{−\frac{\pi}{\mathrm{2}}} \\ $$
Commented by peter frank last updated on 22/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by peter frank last updated on 22/Aug/21
